Answer:
The velocity of the ball after 5 seconds will be 49 m/s
Explanation:
<em>v = final velocity</em>
<em>u = initial velocity</em>
<em>g = acceleration due to gravity</em>
<em>t = time</em>
Initial velocity of the ball = 0 (As the ball is dropped from rest )
Acceleration due to gravity = 9.8 m/s
Time taken = 5 sec
As the acceleration due to gravity is constant in both the cases we can use the equations of motion in order to solve this question
Part I :- As we already know the values of u,g,ant t we can use the first equation of motion in order to find v
Part II :- As we know the values of u, t , g we can use the second equation of motion in order to find s.
Velocity of the ball after 5 seconds
Distance covered by the ball in 5 sec
Answer: In a logical Pace forum subject to the distance
Explanation:
Answer:
A) and B) are correct.
Explanation:
If the object is at rest, it means that no net force is exerted on it.
As the object experiences a downward gravitational force from Earth, in order to be at rest, it must experience an upward force with the same magnitude as the gravitational force on the object.
This force is supplied by the normal force, which can adopt any value in order to meet the condition imposed by Newton´s 2nd Law, and is always perpendicular to the surface on which the object is placed (in this case, the ground).
At a molecular level, this normal force is supplied by the bonded molecules of the ground that behave like small springs being compressed by the molecules of the object, exerting an upward restoring force upward on them.
So, the statements A) and B) are true.
so as this ant moves
5 cm every second you multiply 5 by 120 (60 per minute as there are 60 seconds in a minute)
this is 600 cm
or
6 meters
Answer:
Magnetic flux through the loop is 1.03 T m²
Explanation:
Given:
Magnetic field, B = 4.35 T
Radius of the circular loop, r = 0.280 m
Angle between circular loop and magnetic field, θ = 15.1⁰
Magnetic flux is determine by the relation:
....(1)
Here A represents area of the circular loop.
Area of circular loop, A = πr²
Hence, the equation (1) becomes:

Substitute the suitable values in the above equation.

= 1.03 T m²